97.6k views
0 votes
The density of lithium is 0.530 g/cm^3 What is the volume occupied by 1.20 x 10^3 g of lithium?

User Nick Crews
by
7.0k points

1 Answer

1 vote

Final answer:

The volume occupied by 1.20 x 10^3 g of lithium is 2264.15 cm^3. The volume occupied by 1.20 x 10³ g of lithium is approximately 2264.15 cm³, calculated using the formula volume = mass/density with the given density of 0.530 g/cm³.

Step-by-step explanation:

To find the volume occupied by 1.20 x 10^3 g of lithium, we can use the formula:

Volume = Mass / Density

Substituting the given values:

Volume = 1.20 x 10^3 g / 0.530 g/cm^3

Volume = 2264.15 cm^3

Therefore, the volume occupied by 1.20 x 10^3 g of lithium is 2264.15 cm^3.

The volume occupied by 1.20 x 10³ g of lithium is approximately 2264.15 cm³, calculated using the formula volume = mass/density with the given density of 0.530 g/cm³.

To calculate the volume occupied by a certain mass of lithium, we use the formula for density, which is density = mass/volume. Given the density of lithium as 0.530 g/cm³, and the mass of lithium as 1.20 x 10³ g, we can rearrange the formula to solve for volume: volume = mass/density.

Therefore, the volume occupied by 1.20 x 10³ g of lithium is:

volume = (1.20 x 10³ g) / (0.530 g/cm³) = 2264.15 cm³.

This calculation shows that the volume of lithium that 1.20 x 10³ g occupies is approximately 2264.15 cm³.

User Brandon Rhodes
by
8.6k points